Identifying This Quotient Topology Let $\sim$ relation be on $[0,1]$ as following.
$ x \sim y \Leftrightarrow x=y$ or $x,y \in \{0,1\}$
Show that $[0,1] / \sim$ is homeomorphic to $S=\{x\in \mathbb R^2 : \|x\|=1 \} \subseteq \mathbb R^2$
I need to concrete this quotient set’s elements. What are equivalence classes and what do this set’s elements look like? Also I am not sure on whether I have shown that it is an equivalence relation truly. Showing only one of three conditions of equivalence relation will provide me verifying myself.
It is easy I know but I am confused on this. Thanks for any help
 A: For the homeomorphism part, identify $\mathbb{R}^2$ with the complex plane. 
A natural choice for a homeomorphism is $f:[x]\mapsto e^{2\pi ix}$. This function is surjective because the complex logarithm inverts it on it's image. The injectivity follows from $f([x])=f([y]) \Rightarrow x-y\in\mathbb{Z} \Leftrightarrow x,y \in \left\{0,1\right\}\Leftrightarrow [x]=[y]$.
$f$ is continuous, because the complex exponential function is holomorphic. The continuity of $f^{-1}$ follows from the complex inverse function theorem. 
With kindest regards,
soucerer
A: To verify that $\sim$ is an equivalence relation, you have to check reflexivity, symmetry and transitivity. Note that $x=x$ for all $x\in [0,1]$ and hence $\sim$ is reflexive. Also not that "$x=y$ or $x,y\in\{0,1\}$" is equivalent to "$y=x$ or $y,x\in\{0,1\}$" and hence $\sim$ is symmetric. For transitivity you'll have to do a case distinction and I'll leave that to you.
Given that $\sim$ is indeed an equivalence relation, for $x\in[0,1]$ let us denote by $\overline x \in [0,1]/{\sim}$ the equivalence class of $x$ with respect to $\sim$. For $x\notin\{0,1\}$ we then have singletons $\overline x = \{x\}$ and furthermore $\overline 0 = \overline 1 = \{0,1\}$. Hence we may write
$$
[0,1]/{\sim} = \left.\bigg\{ \,\{x\}\,\middle|\, x\in (0,1)\,\right\} \cup \left.\bigg\{ \{0,1\} \right\}.
$$
A: I will answer in multiple bite size parts (assuming the relation is actually an equivalence relation, proven in another answer on this post):


*

*Explicitly find the elements of $A_1 = [0,1]/ \sim $

*Find a homeomorphism from $A_1$ to $A_2 = [0,1 ) $

*Find a homeomorphism from $A_2$ to $A _3=[0,2\pi ) $

*Find a homeomorphism from  $A_3$ to the unit circle $S $
The equivalence relation states that each $x $ in $(0,1) $ is equivalent to only itself, ie $[x]=x $.   And that $0$ is equivalent to $1$ (so you can choose one of them to be the representative. I choose to denote this by $[0] =0=[1]$). So we have that
$$ A_1 = (0,1) \cup [0] $$
Now we can define the function
$$ f : A_1 \to A_2, \qquad f ([x])= x $$
It is very easily injective and surjective. Also $f $ and its inverse are continuous because the image of open sets are open in both directions.  So $f $ is homeomorphic.  Now define the function
$$ g : A_2 \to A_3, \qquad g (x)=2 \pi x $$
It is very easily injective and surjective. Also $g$ and its inverse are continuous because the image of open sets are open in both directions.  So $g $ is homeomorphic. Finally let us define the function 
$$ h : A_3 \to S, \qquad h (x)= (\cos x , \sin x)$$
It is very easily injective (since the trigomometric functions do not cycle in the range $[0,2\pi) $) and surjective (by definition of trigonometric function). Also $h $ and its inverse are continuous because the image of open sets are open in both directions since the trigonometric functions are continuous.  So $h$ is homeomorphic
Finally we have that $h \circ g \circ f $ is a homeomorpjism from your quotient set to $S $
